Sagot :
It depends on what variable you are tying to solve for first. Say you are trying to solve for x first and then y on the first problem you wrote.
In substitution you solve one of the equations for example with
6x+2y=-10
2x+2y=-10
you solve 2x+2y=-10 for x
2x+2y=-10
-2y = -2y (what you do to one side of the = you do to the other)
2x=-10-2y (to get the variable by its self you divide the # and the variable)
/2=/2 (-10/2=-5 and -2y/2= -y or -1y, they are the same either way)
x=-5-y
now you put that in your original equation that you didn't solve for:
6(-5-y)+2y=-10 solve for that
-30-6y+2y=-10 combine like terms
-30-4y=-10 get the y alone and to do this you first get the -30 away from it
+30=+30
-4y=20 divide the -4 from each side
/-4=/-4 (20/-4=-5)
y=-5
now the equation you previously solved for x can be solved for y.
x=-5-y
x=-5-(-5) a minus parenthesis negative -(- gives you a positive
-5+5=0
x=0
and now we have solved the problem. x=0 and y=-5
In substitution you solve one of the equations for example with
6x+2y=-10
2x+2y=-10
you solve 2x+2y=-10 for x
2x+2y=-10
-2y = -2y (what you do to one side of the = you do to the other)
2x=-10-2y (to get the variable by its self you divide the # and the variable)
/2=/2 (-10/2=-5 and -2y/2= -y or -1y, they are the same either way)
x=-5-y
now you put that in your original equation that you didn't solve for:
6(-5-y)+2y=-10 solve for that
-30-6y+2y=-10 combine like terms
-30-4y=-10 get the y alone and to do this you first get the -30 away from it
+30=+30
-4y=20 divide the -4 from each side
/-4=/-4 (20/-4=-5)
y=-5
now the equation you previously solved for x can be solved for y.
x=-5-y
x=-5-(-5) a minus parenthesis negative -(- gives you a positive
-5+5=0
x=0
and now we have solved the problem. x=0 and y=-5